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\title{第四章：大数定律与中心极限定理}
\author{MSS ET AL}
\date{2018年12月}

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\begin{frame}{内容提要  }

\begin{itemize}

\item  随机变量序列的收敛

\item  依测度收敛，依分布收敛

\item  特征函数的概念

\item  伯努利大数定律，切比雪夫大数定律

\item  中心极限定理的应用

\end{itemize}

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\begin{frame}{本章习题}

\begin{itemize}

\item  (4.1) 1,3,8,12

\item  (4.2) 1,5

\item  (4.3) 1,3,5,7

\item  (4.4) 1,5,9,13,17,21,25

\end{itemize}

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问：设 $X_1, X_2,\cdots, X_n,\cdots $ 和 $X$ 都是随机变量。称序列 $\{X_n\}$ {\color{red}依概率收敛}于 $X$, 是指什么？

答：

\begin{itemize}
\item 概念：$\forall \varepsilon>0, \,\,\, \lim\limits_{n\to\infty} P[|X_n-X|\ge \varepsilon] =0$.
\item 例子：投掷硬币，正面出现的频率 $X_n:=\frac{H_n}{n}$.
\item 伯努利大数定律：频率{\color{red}依概率收敛}于概率。
\end{itemize}


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问：设 $X_1, X_2,\cdots, X_n,\cdots $ 和 $X$ 都是随机变量。设分布函数分别是
$F_1(x)$, $F_2(x)$, $\cdots$, $F_n(x)$, $\cdots$ 和 $F(x)$. 
称序列 $\{X_n\}$ {\color{red}依分布收敛}于 $X$, 是指什么？

答：

\begin{itemize}
\item 对 $F$ 的所有连续点 $x$, 都有 $\lim\limits_{n\to\infty} F_n(x)=F(x)$.
\item 例子：中心极限定理。
\end{itemize}


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问：若两个随机变量序列 $\{X_n\}$ 和 $\{Y_n\}$ 分别依概率收敛于常数 $a$ 和 $b$. 
则随机变量序列 $\{X_n-Y_n\}$ 依概率收敛于常数 $a-b$.

答：

\begin{itemize}
\item $A_n= \{\omega : |(X_n(\omega)-Y_n(\omega))-(a-b)|\ge \epsilon \}$.
\item $B_n= \{\omega : |(X_n(\omega)-a|\ge \epsilon/2 \}$.
\item $C_n= \{\omega : |Y_n(\omega)-b|\ge \epsilon/2 \}$.
\item 由 $A_n\subseteq B_n\cup C_n$ 得证。
\end{itemize}


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证明：若随机变量序列 $\{X_n\}$ {\color{red}依概率收敛}于随机变量 $X$, 则 $\{X_n\}$ 一定{\color{red}依分布收敛}于 $X$.

答：

\begin{enumerate}
\item {\small $\{X\le x'\} = \{X\le x', {\color{red}X_n\le x}\} \cup \{X\le x', {\color{red}X_n> x}\}$.}
\item $F(x')\le F_n(x)+P[|X_n-X|\ge x-x']$.
\item 对任意 $x'<x$, 要证 $F(x')\le \varliminf\limits_{n\to\infty} F_n(x)$.
\item $F(x-0)\le \varliminf\limits_{n\to\infty} F_n(x) \le \varlimsup\limits_{n\to\infty} F_n(x) \le F(x+0)$.
\end{enumerate}



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问：随机变量 $X$ 的{\color{red}特征函数}是什么？%写出计算公式。

答：

\begin{itemize}
\item 特征函数是 $\varphi(t)=E[e^{itX}]$,\,\, $-\infty < t < + \infty$.
\item $\varphi(t) = 1+ it E(X) + \frac{(it)^2}{2!} E(X^2) + \frac{(it)^3}{3!}E(X^3) +\cdots $.
\item 离散型 $\varphi(t)=\sum\limits_{i\in I} e^{itx_k} p_k$.
\item 连续型 $\varphi(t)=\int_{x\in\mathbb{R}} e^{itx}p(x)dx$.
\end{itemize}


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问：证明{\color{red}伯努利大数定律}(LLN)：设 $S_n$ 为 $n$ 重伯努利试验中事件 $A$ 发生的次数。设 $P(A)=p$. 则有
\begin{eqnarray*}
\forall \varepsilon>0,\,\,\,\,\lim\limits_{n\to\infty}  P\left[ \Big{|}\frac{S_n}{n}-p \Big{|}\ge \varepsilon  \right] =0.
\end{eqnarray*}

答：

\begin{itemize}
\item 证明关键是使用切比雪夫不等式。
\item $P\left[ \Big{|}\frac{S_n}{n}-p \Big{|}\ge \varepsilon  \right] \le \frac{Var(S_n/n)}{\varepsilon^2}
=\frac{p(1-p)}{n\varepsilon^2}.$
\end{itemize}


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问：设某硬币出现正反面的概率相等。现投掷1万次，设$\xi$ 为出现正面的频率。估计下述概率：
\begin{eqnarray*}
P \Big{[} |\xi-0.5|\ge 0.01 \Big{]}  \le \,\, ?\,\, {\color{red} (0.25)}
\end{eqnarray*}

答：

\begin{itemize}
\item 这是应用切比雪夫不等式的一个例子。
\item 这里 $n=10000$, $p=0.5$, $\varepsilon=0.01$. 
\item 掷1万次则正面频率在$[0.49,0.51]$的概率超$0.75$.
\end{itemize}


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问：说明用{\color{red}蒙特卡洛方法}计算定积分的思路。

设 $0\le f(x)\le 1$, 其中 $0\le x\le 1$. 计算 $\int_0^1 f(x)dx$.

\begin{itemize}
\item 在 $(x,y)\in [0,1]\times [0,1]$ 单位正方形中任意投点，落在函数图象下方的概率等于该定积分。
\item 计算落在 $y=f(x)$ 的图象下方的频率。
\item 蒙特卡洛方法：用频率估计概率。
\end{itemize}


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问：叙述并证明{\color{red}切比雪夫大数定律}(LLN)。


答：设随机变量 $X_1$, $X_2$, $\cdots$, $X_k$, $\cdots$ 两两不相关。
设它们的数学期望 $\mu_1$, $\mu_2$, $\cdots$, $\mu_k$, $\cdots$ 都存在。 
设它们的方差有公共上界，即 $\exists c$, $\forall k$, $Var(X_k)\le c$. 

记 $Y_n=X_1+\cdots+X_n$. 则有
\begin{eqnarray*}
\forall \varepsilon>0,\,\,\,\,\lim\limits_{n\to\infty}  P\left[ \Big{|} \frac{(Y_n-E(Y_n)}{n}  \Big{|}\ge \varepsilon  \right] =0.
\end{eqnarray*}

证明：使用切比雪夫不等式。


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问：叙述{\color{red}林德伯格-莱维中心极限定理} (CLT)。


答：设 $\{X_n\}$ 是一列独立同分布的随机变量，
设其均值和方差都存在。
%：$E(X_k)=\mu$, $Var(X_k)=\sigma^2$.
记 $Y_n=X_1+\cdots+X_n$. 则有
\begin{itemize}
\item  $Y_n^*:=\frac{Y_n-E(Y_n)}{\sqrt{Var(Y_n)}}$ 近似于服从 $N(0,1)$. 
\item $\lim\limits_{n\to\infty}  P\left[ Y_n^* \le y \right] =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{y} \exp\left(-\frac{t^2}{2}\right)dt$.
\end{itemize}


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问：设 $X_1,X_2,\cdots$ 为独立同分布的随机变量，共同分布为 $U(0,1)$. 记 $Y_n=X_1+\cdots+X_n$. 求 $n\to\infty$ 时，$Y_n$ 的密度函数的演化规律。(这规律称为CLT)

答：


\begin{itemize}
\item 使用卷积公式：{\color{red}独立的随机变量的和}的密度函数是各自密度函数的卷积。
\item $p_2(y)= \int_{x\in\mathbb{R}} p_1(x)p_1(y-x)dx$.
\item $p_3(y)= \int_{x\in\mathbb{R}} p_1(x)p_2(y-x)dx$.
\end{itemize}


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问：某复杂系统由100个独立工作的部件组成，每个部件正常工作的概率是0.9. 求正常工作的部件的数目至少有85个的概率。


\begin{itemize}
\item $X_k=1(0)$ $\Longleftrightarrow$ 第 $k$ 个部件(不)正常工作。
\item 正常工作的部件数目 $Y=X_1+\cdots+X_n$.
\item $n=100$, $P(X_k=1)=0.9$. 要求 $P(Y\ge 85)$.
\end{itemize}


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问：设某药的实际治愈率为0.8.  求100个病患使用此药后至少有75个被治愈的概率。

答：


\begin{itemize}
\item $X_k=1(0)$ $\Longleftrightarrow$ 第 $k$ 个病患(没)被治愈。
\item 被治愈的病患数目 $Y=X_1+\cdots+X_n$.
\item $n=100$, $P(X_k=1)=0.8$. 要求 $P(Y\ge 75)$.
\item {\color{red}由 CLT}, $\frac{Y-E(Y)}{\sqrt{Var(Y)}}$ 近似于标准正态分布。
\end{itemize}


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问：某车间有同型号机床200台独立工作，一小时内每台机床有70\%的时间在工作。每台机床工作时要消耗电能15kW. 为以95\%概率保证车间正常工作，求至少需要多少电能？

答：

\begin{itemize}
\item $X_k=1(0)$ $\Longleftrightarrow$ 第 $k$ 个机床(不)在工作。
\item 在工作的机床数目 $Y=X_1+\cdots+X_n$.
\item $n=200$, $p=0.7$, 求$m$ 使 $P(Y\ge m)=0.95$.
\end{itemize}


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问：为调查某节目在某市的收视率$p$，问至少需要调查多少住户，可以使调查结果 $\hat{p}$ 与真实收视率 $p$ 的差不大于5\%的概率大于等于90\%?

答：

\begin{itemize}
\item $X_k=1(0)$ $\Longleftrightarrow$ 第 $k$ 个住户(不)收看该节目。
\item 收看该节目的住户数目 $Y=X_1+\cdots+X_n$.
\item $\hat{p}= \frac{Y}{n}$. 求 $n$ 使得 $P(|\hat{p}-p|\le 0.05)\ge 0.9$.
\end{itemize}

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